Let $R$ be the region in the first quadrant enclosed by the $y$ -axis, the line $y=1$, the line $y=6$, and the curve $y=\dfrac{1}{x^2}$. $y$ $x$ ${y=\dfrac{1}{x^2}}$ $ R$ $ 1$ $ 6$ A solid is generated by rotating $R$ about the $y$ -axis. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi\int_1^6 \dfrac 1y\,dy$ (Choice B) B $\pi\int_1^6 \dfrac{1}{y^2}\,dy$ (Choice C) C $\pi\int_1^6 \dfrac{1}{y^4}\,dy$ (Choice D) D $\pi\int_1^6 \dfrac{1}{\sqrt y}\,dy$
Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=\dfrac{1}{x^2}}$ Notice the slices are horizontal, because we are rotating $R$ about the $y$ -axis. Each slice is a cylinder. Let the thickness of each slice be $dy$ and let the radius of the base, as a function of $y$, be $r(y)$. Then, the volume of each slice is $\pi [r(y)]^2\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(y)]^2\,dy$ This is called the disc method. What we now need is to figure out the expression of $r(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\dfrac{1}{x^2}}$ $ 1$ $ 6$ $r$ The radius is equal to the distance between the curve $y=\dfrac{1}{x^2}$ and the $y$ -axis. To find it, we need to solve the equation for $x$ : $x=\dfrac{1}{\sqrt y}$ So, for any $y$ -value, $r(y)=\dfrac{1}{\sqrt y}}$. Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(y)}]^2 \\\\ &=\pi\left[\dfrac{1}{\sqrt y}}\right]^2 \\\\ &=\pi \cdot\dfrac 1y \end{aligned}$ The bottom endpoint of $R$ is at $y=1$ and the top endpoint is at $y=6$. So the interval of integration is $[1,6]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_1^6 \pi \cdot\dfrac 1y\,dy \\\\ &=\pi\int_1^6 \dfrac 1y\,dy \end{aligned}$